Problem: Let $f$ be a differentiable function with $f(8)=-5$ and $f'(8)=3$. What is the value of the approximation of $f(8.1)$ using the function's local linear approximation at $x=8$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-4.8$ (Choice B) B $-4.7$ (Choice C) C $-4.6$ (Choice D) D $-4.5$
Answer: The local linear approximation of $f$ at $x=8$ is achieved using the equation of the line tangent to $f$ at $x=8$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=f'(8)(x-8)+f(8)$. Plugging $f(8)=-5$ and $f'(8)=3$, we obtain $L(x)=3(x-8)-5$. To approximate $f(8.1)$, all we need is to plug $x=8.1$ into $L(x)$. $\begin{aligned} L(8.1)&=3(8.1-8)-5 \\\\ &=3(0.1)-5 \\\\ &=-4.7 \end{aligned}$ In conclusion, the approximation of $f(8.1)$ using the function's local linear approximation at $x=8$ is $-4.7$.